The function f(x) = e^x + x,being differentia | vedclass

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(B) Given $f(x) = e^x + x$. We need to find $(f^{-1})'(f(\ln 2))$. By the formula for the derivative of an inverse function,$(f^{-1})'(y) = \frac{1}{f

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$\frac{1}{3}$

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(B) Given $f(x) = e^x + x$. We need to find $(f^{-1})'(f(\ln 2))$. By the formula for the derivative of an inverse function,$(f^{-1})'(y) = \frac{1}{f'(x)}$,where $y = f(x)$. Here,$y = f(\ln 2)$,so $x = \ln 2$. First,find $f'(x)$: $f'(x) = \frac{d}{dx}(e^x + x) = e^x + 1$. Now,evaluate $f'(x)$ at $x = \ln 2$: $f'(\ln 2) = e^{\ln 2} + 1 = 2 + 1 = 3$. Therefore,$(f^{-1})'(f(\ln 2)) = \frac{1}{f'(\ln 2)} = \frac{1}{3}$.

The function $f(x) = e^x + x$,being differentiable and one-to-one,has a differentiable inverse $f^{-1}(x)$. The value of $(f^{-1})'(f(\ln 2))$ is

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